How do you find the equation of the tangent line to the curve #f(x) = sin cos(x)# at x = pi/2?

2 Answers
Apr 28, 2016

For this periodic-wave-function, with period #2pi, (pi/2. 0)# is a tangent-crossing-curve point of inflexion. The equation of the tangent is #x + y = pi/2#.

Explanation:

#f'(x) = - sin x cos (cos x)#.
#f'(pi/2)=-sin (pi/2) cos (cos( pi/2))=(-1)(1)=-1#
#f''(x)=-cos x cos(cos x)-sin x (-sin(cos x)(-sin x)#.
#f''(pi/2)=0#.

So, for this periodic-wave-function, with period #2pi, (pi/2. 0)# is a tangent-crossing-curve point of inflexion.

The slope of the tangent is #-1#.

The equation of the tangent is #x + y = pi/2#.

An idiosyncrasy of this graph is that the amplitude is not 1. It is sin(1 radian)=0.84147, nearly.

This is evident from #|cos x|<=1, and so, |f(x)|<=|sin( 1 radian )|<=0.841471#.

One wave of this periodic graph is given for #x in [0, 2pi]#.

Apr 29, 2016

To find the equation of a line, we need a point and a slope.

#f(x) = sin(cos(x))#

Find the #y# coordinate
of the point on the curve where #x=pi/2#.

#y = f(pi/2) = sin(cos(pi/2)) = sin(0)=0#.

So the point is #(pi/2,0)#

Find the slope
of the tangent line at #x=pi/2#.

Use the derivative. Use the chain rule to differentiate:

#f'(x) = cos(cos(x))(d/dx(cos(x))) = cos(cos(x))(-sin(x))#

At #pi/2#, the slope of the tangent line is

#m = f'(pi/2) = cos(cos(pi/2))(-sin(pi/2))#

# = cos(0)(-1) = (1)(-1) = -1#

Find an equation of the line with slope #m=-1# through the point #(pi/2,0)#

There are various answers possible.

#y-0 = -1(x-pi/2)#

#y = -x+pi/2#

#x+y=pi/2#